# 4.3: Maxwell's Derivation of the Gas-velocity Probability-density Function

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To this point, we have been developing our ability to characterize the gas-velocity distribution functions. We now want to use Maxwell’s argument to find them. We have already introduced the first step, which is the recognition that three-dimensional probability-density functions can be expressed as products of independent one-dimensional functions, and that \({\rho }_{\theta }\left(\theta \right)\), and \({\rho }_{\varphi }\left(\varphi \right)\) are the constants \({1}/{2}\) and \({1}/{2\pi }\). Now, because the probability density associated with any given velocity is just a number that is independent of the coordinate system, we can equate the three-dimensional probability-density functions for Cartesian and spherical coordinates: \(\rho \left(v_x,v_y,v_z\right)=\rho \left(v,\ \theta ,\varphi \right)\) so that

\[{\rho }_x\left(v_x\right){\rho }_y\left(v_y\right){\rho }_z\left(v_z\right)=\frac{{\rho }_v\left(v\right)}{4\pi }\]

We take the partial derivative of this last equation with respect to \(v_x\). The probability densities \({\rho }_y\left(v_y\right)\) and \({\rho }_z\left(v_z\right)\) are independent of \(v_x\). However, \(v\) is a function of \(v_x\), because \(v^2=v^2_x+v^2_y+v^2_z\). We find

\[\frac{{d\rho }_x\left(v_x\right)}{dv_x}{\rho }_y\left(v_y\right){\rho }_z\left(v_z\right)=\frac{1}{4\pi }{\left(\frac{{\partial \rho }_v\left(v\right)}{\partial v_x}\right)}_{v_yv_v}=\frac{1}{4\pi }\left(\frac{d{\rho }_v\left(v\right)}{dv}\right){\left(\frac{\partial v}{\partial v_x}\right)}_{v_yv_z}\] Since \(v^2=v^2_x+v^2_y+v^2_z\), \(2v{\left({\partial v}/{\partial v_x}\right)}_{v_yv_z}=2v_x\) and \[{\left(\frac{\partial v}{\partial v_x}\right)}_{v_yv_z}=\frac{v_x}{v}\]

Making this substitution and dividing by the original equation gives

\[\frac{{d\rho }_x\left(v_x\right)}{dv_x}\frac{{\rho }_y\left(v_y\right){\rho }_z\left(v_z\right)}{{\rho }_x\left(v_x\right){\rho }_y\left(v_y\right){\rho }_z\left(v_z\right)}=\frac{v_x}{v}\frac{1}{{\rho }_v\left(v\right)}\frac{d{\rho }_v\left(v\right)}{dv}\]

Cancellation and rearrangement of the result leads to an equation in which the independent variables \(v_x\) and \(v\) are separated. This means that each term must be equal to a constant, which we take to be \(-\lambda\). We find

\[\left(\frac{1}{v_x \rho_x\left(v_x\right)}\right) \frac{d \rho_x \left(v_x\right)}{dv_x}=\left(\frac{1}{v\rho_v \left(v\right)}\right)\frac{d \rho_v\left(v\right)}{dv}=-\lambda\]

so that

\[\frac{d \rho_x\left(v_x\right)}{ \rho_x\left(v_x\right)}=-\lambda v_x dv_x\]

and

\[\frac{d\rho_v\left(v\right)}{\rho_v\left(v\right)}=-\lambda vdv\]

From the first of these equations, we obtain the probability density function for the distributions of one-dimensional velocities. (See Section 4.4.) The three-dimensional probability density function can be deduced from the one-dimensional function. (See Section 4.5.)

From the second equation, we obtain the three-dimensional probability-density function directly. Integrating from \(v=0\), where \({\rho }_v\left(0\right)\) has a fixed value, to an arbitrary scalar velocity, \(v\), where the scalar-velocity function is \({\rho }_v\left(v\right)\), we have

\[\int^{\rho_v\left(v\right)}_{\rho_v\left(0\right)} \frac{d \rho_v\left(v\right)}{ \rho_v\left(v\right)}=-\lambda \int^v_0 vdv\]

or

\[{\rho }_v\left(v\right)={\rho }_v\left(0\right)exp\left(\frac{-\lambda v^2}{2}\right)\]

The probability-density function for the scalar velocity becomes

\[\frac{df_v\left(v\right)}{dv}=v^2{\rho }_v\left(v\right)={\rho }_v\left(0\right)v^2exp\left(\frac{-\lambda v^2}{2}\right)\]

This is the result we want, except that it contains the unknown parameters \({\rho }_v\left(0\right)\) and \(\lambda\). The value of \({\ \rho }_v\left(0\right)\) must be such as to make the integral over all velocities equal to unity. We require

\[\begin{aligned} 1 & =\int^{\infty }_0 \left(\frac{df_v\left(v\right)}{dv}\right) dv \\ ~ & = \rho_v\left(0\right)\int^{\infty }_0 v^2\mathrm{exp}\left(\frac{-\lambda v^2}{2}\right)dv \\ ~ & =\frac{\rho_v\left(0\right)}{4\pi } \left(\frac{2\pi }{\lambda }\right)^{3/2} \end{aligned}\]

so that

\[\rho_v\left(0\right)=4\pi \left(\frac{\lambda }{2\pi }\right)^{3/2}\]

where we use the definite integral \(\int^{\infty }_0 x^2 \mathrm{exp}\left(-ax^2\right)dx=\left(1/4\right)\sqrt{\pi /a^3}\). (See Appendix D.) The scalar-velocity function in the three-dimensional probability-density function becomes

\[\rho_v\left(v\right)=4\pi \left(\frac{\lambda }{2\pi }\right)^{3/2}\mathrm{exp}\left(\frac{-\lambda v^2}{2}\right)\]

The probability-density function for the scalar velocity becomes

\[\begin{aligned} \frac{df_v\left(v\right)}{dv} & =v^2 \rho_v \left(v\right) \\ ~ & =4\pi \left(\frac{\lambda }{2\pi }\right)^{3/2}v^2\mathrm{exp}\left(\frac{-\lambda v^2}{2}\right) \end{aligned}\]

The three-dimensional probability density in spherical coordinates becomes

\[\begin{aligned} \rho \left(v,\ \theta ,\varphi \right) & = \rho_v\left(v\right)\rho_{\theta}\left(\theta \right) \rho_{\varphi }\left(\varphi \right) \\ ~ & =\left(\frac{\lambda }{2\pi }\right)^{3/2}\mathrm{exp}\left(\frac{-\lambda v^2}{2}\right) \end{aligned}\]

The probability that an arbitrarily selected molecule has a velocity vector whose magnitude lies between \(v\) and \(v+dv\), while its \(\theta\)-component lies between \(\theta\) and\(\ \theta +d\theta\), and its \(\varphi\)-component lies between \(\varphi\) and \(\varphi +d\varphi\) becomes

\[ \begin{aligned} dP\left(\textrm{ʋ}\prime \right) & =\left(\frac{df_v\left(v\right)}{dv}\right)\left(\frac{df_{\theta }\left(\theta \right)}{d\theta }\right)\left(\frac{df_{\varphi }\left(\varphi \right)}{d\varphi }\right)dvd\theta d\varphi \\ ~ & =\rho \left(v,\theta ,\varphi \right)v^2 \mathrm{sin} \theta dvd\theta d\varphi \\ ~ & =\left(\frac{1}{4\pi }\right) \rho_v \left(v\right)v^2 \mathrm{sin} \theta dvd\theta d\varphi \\ ~ & = \left(\frac{\lambda }{2\pi }\right)^{3/2}v^2exp\left(\frac{-\lambda v^2}{2}\right) \mathrm{sin} \theta dvd\theta d\varphi \end{aligned}\]

In Section 4.6, we again derive Boyle’s law and use the ideal gas equation to show that \(\lambda ={m}/{kT}\).